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JulijaS [17]
3 years ago
5

Where on the rope do you expect the tension will be greatest? Where do you expect it will be the least? A uniform rope with leng

th L and mass m is held at one end and whirled in a horizontal circle with angular velocity ω. You can ignore the force of gravity on the rope. (a) At a point on the rope a distance r from the end that is held, what is the tension F? (b) What is the speed of transverse waves at this point? (c) Find the time required for a transverse wave to travel from one end of the rope to the other.
Physics
1 answer:
KonstantinChe [14]3 years ago
7 0

Answer:

a)F=\dfrac{\omega^2 m}{2L}\left({L^2}-{r^2}\right)

b)V=\omega \sqrt{\dfrac{L^2-r^2}{2}}

c)t=\dfrac{\pi}{\omega \sqrt2}

Explanation:

Given that

Length =L

Mass = m

Force on elemental part

dF= dm ω² r

dm = m/L dr

dF= ω² r m/L dr

By integrating from  r to L

F=\dfrac{m}{L} \int_{r}^{L}\omega^2 r \ dr

F=\dfrac{\omega^2 m}{L}\left(\dfrac{L^2}{2}-\dfrac{r^2}{2}\right)

F=\dfrac{\omega^2 m}{2L}\left({L^2}-{r^2}\right)

Velocity V

V=\sqrt{\dfrac{F}{\dfrac{m}{L}}}

V=\sqrt{\dfrac{\dfrac{\omega^2 m}{2L}\left({L^2}-{r^2}\right)}{\dfrac{m}{L}}}

V=\omega \sqrt{\dfrac{L^2-r^2}{2}}

\dfrac{dr}{dt}=V=\omega \sqrt{\dfrac{L^2-r^2}{2}}

\int_{0}^{L}\dfrac{dr}{\sqrt{L^2-r^2}}=\int_{0}^{t} \dfrac{\omega }{\sqrt2}dt

sin^{-1}1=\ \dfrac{\omega }{\sqrt2}t

\dfrac{\pi}{2}=\ \dfrac{\omega }{\sqrt2}t

t=\dfrac{\pi}{\omega \sqrt2}

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Answer:

1.68 s

Explanation:

From newton's equation of motion,

a = (v-u)/t.................................. Equation 1

Making t the subject of the equation

t =(v-u)g............................. Equation 2

Where t = time taken for the bowling pin to reach the maximum height, v = final velocity bowling pin, u = initial velocity of the bowling pin, g = acceleration due to gravity.

Note: Taking upward to be negative and down ward to be positive,

Given: v = 0 m/s ( at the maximum height), u = 8.20 m/s, g = -9.8 m/s²

t = (0-8.20)/-9.8

t = -8.20/-9.8

t = 0.84 s.

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T = 1.68 s.

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Explanation:

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Put the value into the formula

V_{R}=5.10\times(\dfrac{80.0}{\sqrt{(80.0)^2+(\dfrac{1}{2\pi\times\dfrac{1}{2}\times1198.45\times1.66\times10^{-6}})^2}})

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Using formula of  V_{R}

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